The average weight of A, B C is 40 Kg, the average weight of B, D, E is 42 kg and the weight of F is equal to that of B. what is the average weight of A, B, C, D, E and F?

Correct Option (3)

The problem requires calculating the average weight of six individuals (A, B, C, D, E, F) based on given average weights of two groups and a specific weight equivalence.

• Given, the average weight of A, B, C is 40 Kg.
  This implies: $\frac{A+B+C}{3}=40$
  Therefore, $A+B+C=120$ (Equation 1)
• Given, the average weight of B, D, E is 42 Kg.
  This implies: $\frac{B+D+E}{3}=42$
  Therefore, $B+D+E=126$ (Equation 2)
• From Equation 1, we can express B as: $B=120-A-C$
• Substitute this expression for B into Equation 2:
  $(120-A-C)+D+E=126$
  $D+E=126-120+A+C$
  $D+E=6+A+C$ (Equation 3)
• Given, the weight of F is equal to that of B, i.e., $F=B$.
• The average weight of A, B, C, D, E, F is calculated as:
  $\frac{A+B+C+D+E+F}{6}$
• Substitute $F=B$ into the expression:
  $\frac{A+B+C+D+E+B}{6}$
• Now, substitute Equation 1 ($A+B+C=120$) and Equation 3 ($D+E=6+A+C$) into the sum:
  The sum can be written as: $(A+B+C)+(D+E)+B$
  Substitute the known values: $120+(6+A+C)+B$
  Rearrange to group terms: $120+6+(A+C+B)$
  Since $A+C+B=120$ (from Equation 1), the total sum is:
  $120+6+120=246$ Kg
• Finally, calculate the average weight:
  $\frac{246}{6}=41$ Kg

Incorrect Options:

Options 1 (40.5 kg), 2 (40.8 kg), and 4 (Cannot be determined as data inadequate) are incorrect. The provided data is sufficient to determine the average weight of all six individuals through systematic algebraic substitution, leading to a precise value of 41 kg. Therefore, stating that the data is inadequate or arriving at any other numerical value is erroneous.